Question Bank • QB_IDE_EXP_032 NAT • 2 Marks

Section 5: Irrigation and Drainage Engineering

Agricultural Drainage

Subtopic: Planning, design and layout of surface and sub-surface drainage systems

Question Statement:

In Hooghoudt's equation, if the impermeable layer is located at a very large depth below the drain level, the flow becomes radial near the drain and the equivalent depth $d$ approaches a limiting value. For a drain spacing $S = 50\text{ m}$, drain pipe radius $r = 0.10\text{ m}$, and depth to barrier $D = 8.0\text{ m}$, Hooghoudt's equivalent depth is $d = \frac{D}{1 + \frac{8 D}{\pi S} \ln(D/r_0)}$. Taking $\ln(8.0/0.10) = \ln(80) = 4.382$, the equivalent depth $d$ in meters is ________ (round off to 2 decimal places).

Answer Key:

4.47

Step-by-Step Solution & Explanation

Using the formula: $d = \frac{D}{1 + \frac{8 D}{\pi S} \ln(D/r)}$ Given: • $D = 8.0\text{ m}$ • $S = 50.0\text{ m}$ • $r = 0.10\text{ m}$ • $\ln(D/r) = \ln(80) \approx 4.382$ $\frac{8 D}{\pi S} \ln(D/r) = \frac{8 \times 8.0}{\pi \times 50.0} \times 4.382 = \frac{64.0}{157.0796} \times 4.382 = 0.40744 \times 4.382 \approx 1.7854$ $d = \frac{8.0}{1 + 1.7854} = \frac{8.0}{2.7854} \approx 2.872\text{ m}$ Wait, let's recheck: $8.0 / 2.7854 = 2.87\text{ m}$. Setting correct_answer: "2.87", range min: 2.80, max: 2.95.
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